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Magazine Chapter 4: Problem 9
For the curve \(x^{2 / 3}+y^{2 / 3}=4\), find the equations of the tangent lines at \((2 \sqrt{2},-2 \sqrt{2})\), at \((8,0)\), and at \((0,8)\).
Short Answer
Expert verified
The tangent lines are: 1. \( y = -x + 4 \sqrt{2} \) at \( (2 \sqrt{2}, -2 \sqrt{2}) \). 2. \( y = 0 \) at \( (8, 0) \). 3. \( x = 0 \) at \( (0, 8) \).
Step by step solution
01
- Write the given curve equation
The equation of the curve is given by \[ x^{2/3} + y^{2/3} = 4. \]
02
- Differentiate implicitly with respect to x
Differentiate both sides of the equation implicitly with respect to x. Using the chain rule, we get: \[ \frac{2}{3} x^{-1/3} + \frac{2}{3} y^{-1/3} \frac{dy}{dx} = 0. \]
03
- Solve for \( \frac{dy}{dx} \)
Rearrange the differentiated equation to solve for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = - \frac{x^{-1/3}}{y^{-1/3}} = -\frac{y^{1/3}}{x^{1/3}}. \]
04
- Find the slope at \( (2 \sqrt{2}, -2 \sqrt{2}) \)
Plug in \( x = 2 \sqrt{2} \) and \( y = -2 \sqrt{2} \) into the slope formula: \[ \frac{dy}{dx} \bigg|_{(2 \sqrt{2}, -2 \sqrt{2})} = -\frac{(-2 \sqrt{2})^{1/3}}{(2 \sqrt{2})^{1/3}}. \] Since \((2 \sqrt{2})^{1/3} = 2^{1/3} \cdot 2^{1/3} = 2^{2/3}\), thus \( \frac{dy}{dx} \bigg|_{(2 \sqrt{2}, -2 \sqrt{2})} = -1. \)
05
- Write the tangent line equation at \( (2 \sqrt{2}, -2 \sqrt{2}) \)
Using the point-slope form of the line equation: \[ y - y_1 = m (x - x_1), \] where \( x_1 = 2 \sqrt{2} \), \( y_1 = -2 \sqrt{2} \), and \(m = -1\), we get: \[ y + 2 \sqrt{2} = -1 (x - 2 \sqrt{2}). \] Simplifying, the equation is: \[ y = -x + 4 \sqrt{2}. \]
06
- Find the slope at \( (8, 0) \)
Plug in \( x = 8 \) and \( y = 0 \) into the slope formula: \[ \frac{dy}{dx} \bigg|_{(8, 0)} = -\frac{0^{1/3}}{8^{1/3}} = 0. \]
07
- Write the tangent line equation at \( (8, 0) \)
Using the point-slope form: \[ y - 0 = 0 (x - 8), \] Thus, the equation is: \[ y = 0. \]
08
- Find the slope at \( (0, 8) \)
Plug in \( x = 0 \) and \( y = 8 \) into the slope formula: \[ \frac{dy}{dx} \bigg|_{(0, 8)} = - \frac{8^{1/3}}{0^{1/3}}. \] Since division by zero is undefined, the slope is vertical.
09
- Write the tangent line equation at \( (0, 8) \)
For a vertical line passing through \( x = 0 \): The equation is: \[ x = 0. \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Tangent Line Equations
To understand tangent line equations, we need to first grasp what a tangent line is. Imagine a curve in a graph. A tangent line touches this curve at exactly one point. This line has the same slope as the curve at that point.
The equation for a tangent line can be derived using the point-slope form formula: $$ y - y_1 = m(x - x_1), $$ where \x_1, y_1\ is the point of tangency and m is the slope of the tangent line.
For example, for the curve \( x^{2/3} + y^{2/3} = 4 \), at the point \(2\text{鈭殅2, -2\text{鈭殅2\), we found that the slope m is -1. Thus, the tangent line equation becomes $$ y + 2 \text{鈭殅 2 = -1(x - 2 \text{鈭殅 2). $$ When simplified, this gives us: $$ y = -x + 4 \text{鈭殅 2. $$
The equation for a tangent line can be derived using the point-slope form formula: $$ y - y_1 = m(x - x_1), $$ where \x_1, y_1\ is the point of tangency and m is the slope of the tangent line.
For example, for the curve \( x^{2/3} + y^{2/3} = 4 \), at the point \(2\text{鈭殅2, -2\text{鈭殅2\), we found that the slope m is -1. Thus, the tangent line equation becomes $$ y + 2 \text{鈭殅 2 = -1(x - 2 \text{鈭殅 2). $$ When simplified, this gives us: $$ y = -x + 4 \text{鈭殅 2. $$
Slope Calculation
The slope of a function at a point tells us how steep the curve is at that point. It's a measure of the rate of change of the function.
To find the slope of the curve \(x^{2/3} + y^{2/3} = 4 \), we use implicit differentiation. Differentiating both sides with respect to x, we have: $$ \frac{2}{3} x^{-1/3} + \frac{2}{3} y^{-1/3} \frac{dy}{dx} = 0. $$ Solving for \(\frac{dy}{dx}\): $$ \frac{dy}{dx} = - \frac{x^{-1/3}}{y^{-1/3}} = -\frac{y^{1/3}}{x^{1/3}}. $$
This formula allows us to plug in coordinates of specific points to find the slope. For instance, at \( 8, 0 \), plugging the values into our slope formula gives us: $$ \frac{dy}{dx} \bigg|_{(8, 0)} = -\frac{0^{1/3}}{8^{1/3}} = 0. $$ As a result, the slope here is 0, indicating a horizontal tangent line.
To find the slope of the curve \(x^{2/3} + y^{2/3} = 4 \), we use implicit differentiation. Differentiating both sides with respect to x, we have: $$ \frac{2}{3} x^{-1/3} + \frac{2}{3} y^{-1/3} \frac{dy}{dx} = 0. $$ Solving for \(\frac{dy}{dx}\): $$ \frac{dy}{dx} = - \frac{x^{-1/3}}{y^{-1/3}} = -\frac{y^{1/3}}{x^{1/3}}. $$
This formula allows us to plug in coordinates of specific points to find the slope. For instance, at \( 8, 0 \), plugging the values into our slope formula gives us: $$ \frac{dy}{dx} \bigg|_{(8, 0)} = -\frac{0^{1/3}}{8^{1/3}} = 0. $$ As a result, the slope here is 0, indicating a horizontal tangent line.
Curve Analysis
To fully understand a curve, we look at its equation, slopes, and behavior at various points.
The given curve is \( x^{2/3} + y^{2/3} = 4 \). This curve forms a specific shape, and we can analyze it by looking at tangent lines and slopes at key points.
Let's examine the point \(0, 8\). When we use our slope formula: $$ \frac{dy}{dx} \bigg|_{(0, 8)} = - \frac{8^{1/3}}{0^{1/3}}. $$ Since division by zero is undefined, the slope here is vertical. This means that the tangent line is a vertical line passing through \(0, 8\), thus: $$ x = 0. $$
These methods help us better understand the behavior of functions and their graphical representations, aiding us in solving more complex calculus problems.
The given curve is \( x^{2/3} + y^{2/3} = 4 \). This curve forms a specific shape, and we can analyze it by looking at tangent lines and slopes at key points.
Let's examine the point \(0, 8\). When we use our slope formula: $$ \frac{dy}{dx} \bigg|_{(0, 8)} = - \frac{8^{1/3}}{0^{1/3}}. $$ Since division by zero is undefined, the slope here is vertical. This means that the tangent line is a vertical line passing through \(0, 8\), thus: $$ x = 0. $$
These methods help us better understand the behavior of functions and their graphical representations, aiding us in solving more complex calculus problems.
